$(3x^2 +6y)dx -14yzdy +20xz^2$ is an exact differential then why the curl of $(3x^2+6y) \hat{i} -14yz\hat{j} +20xz^2\hat{k} $ is not zero?

Question

$(3x^2 +6y)dx -14yzdy +20xz^2$ is an exact differential then why the curl of $(3x^2+6y) \hat{i} -14yz\hat{j} +20xz^2\hat{k} $ is not zero ? If you experiment if the differential equation $$\partial ^2 {(3x^2 +6y)} / \partial y \partial z$$ = $$\partial ^2 {-14yz}/\partial z\partial x$$= $$\partial ^2 {20xz^2} /\partial x \partial y$$=0 but the curl of $$(3x^2+6y) \hat{i} -14yz\hat{j} +20xz^2\hat{k} $$ is not zero. Now if the differential equation is exact then the vector should be a gradient whose curl should be zero . Where am I wrong?

Answer 1

The curl equals $(14y, -20 z^2, -6) =\left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \mathbf{i} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \mathbf{j} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \mathbf{k} = \begin{bmatrix}\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \\ \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \\ \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\end{bmatrix}$ There is no second order differential unlike what you posted.